can retrieve the correct data

[williamh69]

hi guys i have this two db's

 

menu                       paginas

 

menu_id                  page_id

name                       menu_id

                                title

                                content

 

I populate the menu with the following code, which working fine.

 <?php
function query($parent_id) { //function to run a query
	$query = mysql_query ( "SELECT * FROM menus" );
	return $query;
}

function fetch_menu($query) {
	while ( $result = mysql_fetch_array ( $query ) ) {
		$menu_id = $result ['menu_id'];
		$menu_name = $result ['menu_name'];
		$menu_link = $result ['menu_link'];
		echo "<li  class='has-sub '><a href='index.php?content=paginas&cat=$menu_id'><span>$menu_name</span></a>";

		echo "</li>";
 }
}
fetch_menu (query(0)); //call this function with 0 parent id
?>

then I link  every menu in my list with each page with this code, but appears the first entry in all the menus, I used the following code:

<?php

         $menu_id = $_GET['cat'];

         $query="SELECT  page_id, title, sub_title, content from paginas where menu_id=$page_id";

         $result=mysql_query($query);

         while($row=mysql_fetch_array($result,MYSQL_ASSOC))

         {

                  $page_id = $row['page_id'];
                  $title = $row['title'];
                  $sub_title = $row['sub_title'];
                  $content = $row['content'];



             echo "<h1>$title</h1>";
             echo"<br>";
             echo"<h2>$sub_title</h2>";
             echo"<br>";
             echo"<p>$content</p>";


         }



?>

what i am doing wrong..... thank you for ur help

[adam_bray]

Where is $page_id being set?

 

Also, you can use a MySQL JOIN to cut the query count to 1 -

<?php

	if($stmt = $mysqli->prepare(' SELECT paginas.page_id
	, paginas.title
	, paginas.sub_title
	, paginas.content
	, menus.name AS menu_name
	, menus.menus_link 
	FROM paginas
	LEFT OUTER
	  JOIN menus
	  	ON menus.menu_id = paginas.menu_id 
	WHERE paginas.menu_id=? '))
	{
		$stmt->bind_param("s", $_GET['cat']);
	
		$stmt->execute();	
		$result = $stmt->get_result();

		while( $row = $result->fetch_assoc() )
		{		
		    // Page data
            $page_id 	= $row['page_id'];
           	    $title 		= $row['title'];
          	    $sub_title 	= $row['sub_title'];
       		    $content 	= $row['content'];
			
			// Menu data
			$menu_name	= $row['menu_name'];
			$menu_link	= $row['menu_link'];
		}
		
		echo '<h1>'.$title.'</h1>
		<h2>'.$sub_title.'</h2>
		<p>'.$content.'</p>';

		
		$stmt->close();
	}

?>

[cyberRobot]

The GET variable from the links is being assigned to $menu_id. But your query is using a variable called $page_id. Have you tried the following:

$menu_id = $_GET['cat'];
 
$query="SELECT  page_id, title, sub_title, content from paginas where menu_id=$menu_id";

[williamh69]

hi cyber i tried but t it gives me the same result

[williamh69]

adam i used your code but gives me also the same result

[ginerjm]

Show us the NEW code that you are using.

[williamh69]

<?php
         $menu_id = $_GET['cat'];

         $query="SELECT  * from paginas WHERE menu_id=$menu_id";

         $result=mysql_query($query);

         while($row=mysql_fetch_array($result,MYSQL_ASSOC))

         {

                  $page_id = $row['page_id'];
                  $title = $row['title'];
                  $sub_title = $row['sub_title'];
                  $content = $row['content'];



             echo "<h1>$title</h1>";
             echo"<br>";
             echo"<h2>$sub_title</h2>";
             echo"<br>";
             echo"<p>$content</p>";


         }



?>

here it is

[ginerjm]

Add a test of your query execution to be sure it's a valid query (it has to be bad and not performing and therefore none of your follow up code is even attempted). 

 

Tip - One should ALWAYS check query results and any other dramatic function call that affects the proper execution of your script.